When does every compact $A$ in a topology satisfy $A⊆B⊆C$ for some open $B$ and compact $C$?

I don't know if the property has a name, but it doesn't have to hold even for metric spaces. Consider $X=\mathbb{Q}$ with the Euclidean topology and let $A=\{0\}$. Clearly no open subset of $\mathbb{Q}$ is relatively compact (which is the same claim for metric, or even Hausdorff, spaces).

Lemma. If $X$ is Hausdorff then the property is equivalent to local compactness.

Proof. Note that for Hausdorff spaces the property "$V$ is contained in a compact set" is equivalent to "$\overline{V}$ is compact" which is also know as "$V$ is relatively compact".

"$\Rightarrow$" Since $\{x\}$ is compact then by our property it has open neighbourhood $U$ such that $\overline{U}$ is compact. Hence local compactness.

"$\Leftarrow$" Let $A\subseteq X$ be compact. Then for any $x\in A$ there is an open neighbourhood $U_x\subseteq X$ of $x$ that is relatively compact. Since $\{U_x\}_{x\in A}$ cover $A$ then by compactness $A$ is covered by $U_{x_1},\ldots,U_{x_n}$. Clearly $U_{x_1}\cup\cdots\cup U_{x_n}$ is the neighbourhood we are looking for. $\Box$

For non-Hausdorff spaces I suppose we can treat the property as one of the many non-equivalent definitions of local compactness. The name looks appropriate.

This condition is a stronger version of one form of local compactness (which considers the case that $A$ is a point); maybe it's equivalent to local compactness under some mild hypotheses, I don't know. In any case it implies local compactness (at least if we also assume that the space is Hausdorff so that all the usual definitions are equivalent), so any non-locally compact space is a counterexample and these are plentiful.

For an explicit counterexample consider any infinite-dimensional normed vector space in the norm topology. By Riesz's lemma we know that the closed unit ball is not compact; this implies that no open subset is contained in a compact subset.